Part 1
What is a force?
Force and gravity
Other forces
Forces and vectors
Part 2
Forces and equilibrium
Kinematics
Position Vectors
Constant acceleration
Part 3
Newton’s first law
Newton’s second law
Newton’s third law
Mathematical modelling
Practice tests
Practice Test Paper 1
Solution to Test Paper 1
Practice Test Paper 2
Solution to Test Paper 2
Practice Test Paper 3
Solution to Test Paper 3
Part 4
Moments
Equilibrium
Momentum and impulse
Conservation of momentum
Part 5
Conservation of energy
Variable acceleration I
Variable acceleration II
Circular Motion
Search
Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing
button on the jsMath control panel.
jsMath
Warning:
jsMath
requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.
Solution 19.5a
From Mechanics
Revision as of 15:37, 11 October 2010 by
Ian
(
Talk
|
contribs
)
(diff) ←Older revision |
Current revision
(
diff
) |
Newer revision→
(
diff
)
Jump to:
navigation
,
search
v
=
adt
=
−
kt
d
t
=
−
2
k
t
2
+
c
t
=
0
v
=
2
0
c
=
2
0
v
=
2
0
−
2
k
t
2
Using
t
=
4
0
v
=
5
gives
5
=
2
0
−
5
k
40
2
320
k
=
1
5
k
=
15
320
=
3
64
